The condition on a and b, such that the portion of the line ax + by - 1 = 0 intercepted between the lines a x + y = 0 and x + by = 0 subtends a right angle at the origin, is

Question

The condition on a and b, such that the portion of the line ax + by - 1 = 0 intercepted between the lines a x + y = 0 and x + by = 0 subtends a right angle at the origin, is (A) a = b (B) a + b = 0 (C) a = 2b (D) 2a = b

Tardigrade Admin · Accepted Answer

Lines a x + y = 0 and x + b y = 0 intersect at O (0 , 0). <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/76109e3ac5f6431637727201d1c0cdf8-.png" /> Hence, if A B subtends right angle at O (0 , 0), then ax + y = 0 and x + by = 0 are perpendicular to each other. So, (-a)(-(1/b)) = -1 or a + b = 0

Q. The condition on $a$ and $b$ , such that the portion of the line $a x + b y - 1 = 0$ intercepted between the lines $a x + y = 0$ and $x + b y = 0$ subtends a right angle at the origin, is

$a = b$

$a + b = 0$

$a = 2 b$

$2 a = b$

Lines $a x + y = 0$ and $x + b y = 0$ intersect at $O (0, 0)$ .

Hence, if $A B$ subtends right angle at $O (0, 0)$ , then $a x + y = 0$ and $x + b y = 0$ are perpendicular to each other. So,
$(- a) (- \frac{1}{b}) = - 1$
or $a + b = 0$

Q. The condition on a and b, such that the portion of the line ax+by−1=0 intercepted between the lines ax+y=0 and x+by=0 subtends a right angle at the origin, is

a=b

a+b=0

a=2b

2a=b